3.58 \(\int \frac{\tan ^2(x)}{(a+b \cot ^2(x))^{5/2}} \, dx\)

Optimal. Leaf size=141 \[ \frac{(a-4 b) (3 a-2 b) \tan (x) \sqrt{a+b \cot ^2(x)}}{3 a^3 (a-b)^2}+\frac{b (7 a-4 b) \tan (x)}{3 a^2 (a-b)^2 \sqrt{a+b \cot ^2(x)}}+\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \cot (x)}{\sqrt{a+b \cot ^2(x)}}\right )}{(a-b)^{5/2}}+\frac{b \tan (x)}{3 a (a-b) \left (a+b \cot ^2(x)\right )^{3/2}} \]

[Out]

ArcTan[(Sqrt[a - b]*Cot[x])/Sqrt[a + b*Cot[x]^2]]/(a - b)^(5/2) + (b*Tan[x])/(3*a*(a - b)*(a + b*Cot[x]^2)^(3/
2)) + ((7*a - 4*b)*b*Tan[x])/(3*a^2*(a - b)^2*Sqrt[a + b*Cot[x]^2]) + ((a - 4*b)*(3*a - 2*b)*Sqrt[a + b*Cot[x]
^2]*Tan[x])/(3*a^3*(a - b)^2)

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Rubi [A]  time = 0.241336, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.412, Rules used = {3670, 472, 579, 583, 12, 377, 203} \[ \frac{(a-4 b) (3 a-2 b) \tan (x) \sqrt{a+b \cot ^2(x)}}{3 a^3 (a-b)^2}+\frac{b (7 a-4 b) \tan (x)}{3 a^2 (a-b)^2 \sqrt{a+b \cot ^2(x)}}+\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \cot (x)}{\sqrt{a+b \cot ^2(x)}}\right )}{(a-b)^{5/2}}+\frac{b \tan (x)}{3 a (a-b) \left (a+b \cot ^2(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]^2/(a + b*Cot[x]^2)^(5/2),x]

[Out]

ArcTan[(Sqrt[a - b]*Cot[x])/Sqrt[a + b*Cot[x]^2]]/(a - b)^(5/2) + (b*Tan[x])/(3*a*(a - b)*(a + b*Cot[x]^2)^(3/
2)) + ((7*a - 4*b)*b*Tan[x])/(3*a^2*(a - b)^2*Sqrt[a + b*Cot[x]^2]) + ((a - 4*b)*(3*a - 2*b)*Sqrt[a + b*Cot[x]
^2]*Tan[x])/(3*a^3*(a - b)^2)

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rule 472

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*(e*x
)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*e*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d)*(
p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*(b*c - a*d)*(p + 1) + d*b*(m + n*(
p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p
, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 579

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> -Simp[((b*e - a*f)*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*g*n*(b*c - a*d)*(p +
1)), x] + Dist[1/(a*n*(b*c - a*d)*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f)*(
m + 1) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
e, f, g, m, q}, x] && IGtQ[n, 0] && LtQ[p, -1]

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\tan ^2(x)}{\left (a+b \cot ^2(x)\right )^{5/2}} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\cot (x)\right )\\ &=\frac{b \tan (x)}{3 a (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{3 a-4 b-4 b x^2}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\cot (x)\right )}{3 a (a-b)}\\ &=\frac{b \tan (x)}{3 a (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}+\frac{(7 a-4 b) b \tan (x)}{3 a^2 (a-b)^2 \sqrt{a+b \cot ^2(x)}}-\frac{\operatorname{Subst}\left (\int \frac{(a-4 b) (3 a-2 b)-2 (7 a-4 b) b x^2}{x^2 \left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\cot (x)\right )}{3 a^2 (a-b)^2}\\ &=\frac{b \tan (x)}{3 a (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}+\frac{(7 a-4 b) b \tan (x)}{3 a^2 (a-b)^2 \sqrt{a+b \cot ^2(x)}}+\frac{(a-4 b) (3 a-2 b) \sqrt{a+b \cot ^2(x)} \tan (x)}{3 a^3 (a-b)^2}+\frac{\operatorname{Subst}\left (\int \frac{3 a^3}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\cot (x)\right )}{3 a^3 (a-b)^2}\\ &=\frac{b \tan (x)}{3 a (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}+\frac{(7 a-4 b) b \tan (x)}{3 a^2 (a-b)^2 \sqrt{a+b \cot ^2(x)}}+\frac{(a-4 b) (3 a-2 b) \sqrt{a+b \cot ^2(x)} \tan (x)}{3 a^3 (a-b)^2}+\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\cot (x)\right )}{(a-b)^2}\\ &=\frac{b \tan (x)}{3 a (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}+\frac{(7 a-4 b) b \tan (x)}{3 a^2 (a-b)^2 \sqrt{a+b \cot ^2(x)}}+\frac{(a-4 b) (3 a-2 b) \sqrt{a+b \cot ^2(x)} \tan (x)}{3 a^3 (a-b)^2}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-(-a+b) x^2} \, dx,x,\frac{\cot (x)}{\sqrt{a+b \cot ^2(x)}}\right )}{(a-b)^2}\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \cot (x)}{\sqrt{a+b \cot ^2(x)}}\right )}{(a-b)^{5/2}}+\frac{b \tan (x)}{3 a (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}+\frac{(7 a-4 b) b \tan (x)}{3 a^2 (a-b)^2 \sqrt{a+b \cot ^2(x)}}+\frac{(a-4 b) (3 a-2 b) \sqrt{a+b \cot ^2(x)} \tan (x)}{3 a^3 (a-b)^2}\\ \end{align*}

Mathematica [C]  time = 8.0576, size = 1450, normalized size = 10.28 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Tan[x]^2/(a + b*Cot[x]^2)^(5/2),x]

[Out]

(Sin[x]^2*((-16*b^3*(Cot[x] + Cot[x]^3)^2)/(a*(a - b)^2) + (40*b*Csc[x]^2)/(a - b) + (160*b^2*Cot[x]^2*Csc[x]^
2)/(3*a*(a - b)) + (64*b^3*Cot[x]^4*Csc[x]^2)/(3*a^2*(a - b)) - (40*b^2*Csc[x]^4)/(a - b)^2 + (92*(a - b)*Cos[
x]^2*Hypergeometric2F1[2, 2, 9/2, ((a - b)*Cos[x]^2)/a])/(105*a) + (124*(a - b)*b*Cos[x]^2*Cot[x]^2*Hypergeome
tric2F1[2, 2, 9/2, ((a - b)*Cos[x]^2)/a])/(35*a^2) + (152*(a - b)*b^2*Cos[x]^2*Cot[x]^4*Hypergeometric2F1[2, 2
, 9/2, ((a - b)*Cos[x]^2)/a])/(35*a^3) + (176*(a - b)*b^3*Cos[x]^2*Cot[x]^6*Hypergeometric2F1[2, 2, 9/2, ((a -
b)*Cos[x]^2)/a])/(105*a^4) + (24*(a - b)*Cos[x]^2*HypergeometricPFQ[{2, 2, 2}, {1, 9/2}, ((a - b)*Cos[x]^2)/a
])/(35*a) + (16*(a - b)*b*Cos[x]^2*Cot[x]^2*HypergeometricPFQ[{2, 2, 2}, {1, 9/2}, ((a - b)*Cos[x]^2)/a])/(7*a
^2) + (88*(a - b)*b^2*Cos[x]^2*Cot[x]^4*HypergeometricPFQ[{2, 2, 2}, {1, 9/2}, ((a - b)*Cos[x]^2)/a])/(35*a^3)
+ (32*(a - b)*b^3*Cos[x]^2*Cot[x]^6*HypergeometricPFQ[{2, 2, 2}, {1, 9/2}, ((a - b)*Cos[x]^2)/a])/(35*a^4) +
(16*(a - b)*Cos[x]^2*HypergeometricPFQ[{2, 2, 2, 2}, {1, 1, 9/2}, ((a - b)*Cos[x]^2)/a])/(105*a) + (16*(a - b)
*b*Cos[x]^2*Cot[x]^2*HypergeometricPFQ[{2, 2, 2, 2}, {1, 1, 9/2}, ((a - b)*Cos[x]^2)/a])/(35*a^2) + (16*(a - b
)*b^2*Cos[x]^2*Cot[x]^4*HypergeometricPFQ[{2, 2, 2, 2}, {1, 1, 9/2}, ((a - b)*Cos[x]^2)/a])/(35*a^3) + (16*(a
- b)*b^3*Cos[x]^2*Cot[x]^6*HypergeometricPFQ[{2, 2, 2, 2}, {1, 1, 9/2}, ((a - b)*Cos[x]^2)/a])/(105*a^4) + (20
*a*Sec[x]^2)/(3*(a - b)) - (30*a*b*Csc[x]^2*Sec[x]^2)/(a - b)^2 - (5*a^2*Sec[x]^4)/(a - b)^2 + (5*ArcSin[Sqrt[
((a - b)*Cos[x]^2)/a]])/((((a - b)*Cos[x]^2)/a)^(5/2)*Sqrt[((a + b*Cot[x]^2)*Sin[x]^2)/a]) + (30*b*ArcSin[Sqrt
[((a - b)*Cos[x]^2)/a]]*Cot[x]^2)/(a*(((a - b)*Cos[x]^2)/a)^(5/2)*Sqrt[((a + b*Cot[x]^2)*Sin[x]^2)/a]) + (40*b
^2*ArcSin[Sqrt[((a - b)*Cos[x]^2)/a]]*Cot[x]^4)/(a^2*(((a - b)*Cos[x]^2)/a)^(5/2)*Sqrt[((a + b*Cot[x]^2)*Sin[x
]^2)/a]) + (16*b^3*ArcSin[Sqrt[((a - b)*Cos[x]^2)/a]]*Cot[x]^6)/(a^3*(((a - b)*Cos[x]^2)/a)^(5/2)*Sqrt[((a + b
*Cot[x]^2)*Sin[x]^2)/a]) + (5*ArcSin[Sqrt[((a - b)*Cos[x]^2)/a]])/Sqrt[((a - b)*Cos[x]^2*(a + b*Cot[x]^2)*Sin[
x]^2)/a^2] + (30*b*ArcSin[Sqrt[((a - b)*Cos[x]^2)/a]]*Cot[x]^2)/(a*Sqrt[((a - b)*Cos[x]^2*(a + b*Cot[x]^2)*Sin
[x]^2)/a^2]) + (40*b^2*ArcSin[Sqrt[((a - b)*Cos[x]^2)/a]]*Cot[x]^4)/(a^2*Sqrt[((a - b)*Cos[x]^2*(a + b*Cot[x]^
2)*Sin[x]^2)/a^2]) + (16*b^3*ArcSin[Sqrt[((a - b)*Cos[x]^2)/a]]*Cot[x]^6)/(a^3*Sqrt[((a - b)*Cos[x]^2*(a + b*C
ot[x]^2)*Sin[x]^2)/a^2]) - (60*b*ArcSin[Sqrt[((a - b)*Cos[x]^2)/a]]*Csc[x]^2)/((a - b)*Sqrt[((a - b)*Cos[x]^2*
(a + b*Cot[x]^2)*Sin[x]^2)/a^2]) - (80*b^2*ArcSin[Sqrt[((a - b)*Cos[x]^2)/a]]*Cot[x]^2*Csc[x]^2)/(a*(a - b)*Sq
rt[((a - b)*Cos[x]^2*(a + b*Cot[x]^2)*Sin[x]^2)/a^2]) - (32*b^3*ArcSin[Sqrt[((a - b)*Cos[x]^2)/a]]*Cot[x]^4*Cs
c[x]^2)/(a^2*(a - b)*Sqrt[((a - b)*Cos[x]^2*(a + b*Cot[x]^2)*Sin[x]^2)/a^2]) - (10*a*ArcSin[Sqrt[((a - b)*Cos[
x]^2)/a]]*Sec[x]^2)/((a - b)*Sqrt[((a - b)*Cos[x]^2*(a + b*Cot[x]^2)*Sin[x]^2)/a^2]))*Tan[x])/(a^2*Sqrt[a + b*
Cot[x]^2]*(1 + (b*Cot[x]^2)/a))

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Maple [B]  time = 0.48, size = 1040, normalized size = 7.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)^2/(a+b*cot(x)^2)^(5/2),x)

[Out]

-1/3*b^2/a^3/((a*(a-b))^(1/2)-a+b)^2/((a*(a-b))^(1/2)+a-b)^2/(-a+b)^(1/2)*(-1+cos(x))^2*(cos(x)+1)^2*(cos(x)^2
*a-b*cos(x)^2-a)*(-3*cos(x)^4*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)*ln(4*cos(x)*(-a+b)^(1/2)*(-(cos(
x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)-4*a*cos(x)+4*b*cos(x)+4*(-a+b)^(1/2)*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos
(x)+1)^2)^(1/2))*a^4+3*cos(x)^4*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)*ln(4*cos(x)*(-a+b)^(1/2)*(-(co
s(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)-4*a*cos(x)+4*b*cos(x)+4*(-a+b)^(1/2)*(-(cos(x)^2*a-b*cos(x)^2-a)/(c
os(x)+1)^2)^(1/2))*a^3*b+3*cos(x)^4*(-a+b)^(1/2)*a^4-12*cos(x)^4*(-a+b)^(1/2)*a^3*b+27*cos(x)^4*(-a+b)^(1/2)*a
^2*b^2-26*cos(x)^4*(-a+b)^(1/2)*a*b^3+8*cos(x)^4*(-a+b)^(1/2)*b^4-3*cos(x)^3*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(
x)+1)^2)^(1/2)*ln(4*cos(x)*(-a+b)^(1/2)*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)-4*a*cos(x)+4*b*cos(x)+
4*(-a+b)^(1/2)*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2))*a^4+3*cos(x)^3*(-(cos(x)^2*a-b*cos(x)^2-a)/(co
s(x)+1)^2)^(1/2)*ln(4*cos(x)*(-a+b)^(1/2)*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)-4*a*cos(x)+4*b*cos(x
)+4*(-a+b)^(1/2)*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2))*a^3*b+3*cos(x)^2*(-(cos(x)^2*a-b*cos(x)^2-a)
/(cos(x)+1)^2)^(1/2)*ln(4*cos(x)*(-a+b)^(1/2)*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)-4*a*cos(x)+4*b*c
os(x)+4*(-a+b)^(1/2)*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2))*a^4-6*cos(x)^2*(-a+b)^(1/2)*a^4+18*cos(x
)^2*(-a+b)^(1/2)*a^3*b-27*cos(x)^2*(-a+b)^(1/2)*a^2*b^2+12*cos(x)^2*(-a+b)^(1/2)*a*b^3+3*cos(x)*(-(cos(x)^2*a-
b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)*ln(4*cos(x)*(-a+b)^(1/2)*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)-4*a
*cos(x)+4*b*cos(x)+4*(-a+b)^(1/2)*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2))*a^4+3*(-a+b)^(1/2)*a^4-6*(-
a+b)^(1/2)*a^3*b+3*(-a+b)^(1/2)*a^2*b^2)/cos(x)/((cos(x)^2*a-b*cos(x)^2-a)/(cos(x)^2-1))^(5/2)/sin(x)^9

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^2/(a+b*cot(x)^2)^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 2.99412, size = 1477, normalized size = 10.48 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^2/(a+b*cot(x)^2)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*(a^5*tan(x)^4 + 2*a^4*b*tan(x)^2 + a^3*b^2)*sqrt(-a + b)*log(-(a^2*tan(x)^4 - 2*(3*a^2 - 4*a*b)*tan(
x)^2 + a^2 - 8*a*b + 8*b^2 + 4*(a*tan(x)^3 - (a - 2*b)*tan(x))*sqrt(-a + b)*sqrt((a*tan(x)^2 + b)/tan(x)^2))/(
tan(x)^4 + 2*tan(x)^2 + 1)) - 4*(3*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*tan(x)^5 + 3*(2*a^4*b - 9*a^3*b^2 + 1
1*a^2*b^3 - 4*a*b^4)*tan(x)^3 + (3*a^3*b^2 - 17*a^2*b^3 + 22*a*b^4 - 8*b^5)*tan(x))*sqrt((a*tan(x)^2 + b)/tan(
x)^2))/(a^6*b^2 - 3*a^5*b^3 + 3*a^4*b^4 - a^3*b^5 + (a^8 - 3*a^7*b + 3*a^6*b^2 - a^5*b^3)*tan(x)^4 + 2*(a^7*b
- 3*a^6*b^2 + 3*a^5*b^3 - a^4*b^4)*tan(x)^2), 1/6*(3*(a^5*tan(x)^4 + 2*a^4*b*tan(x)^2 + a^3*b^2)*sqrt(a - b)*a
rctan(2*sqrt(a - b)*sqrt((a*tan(x)^2 + b)/tan(x)^2)*tan(x)/(a*tan(x)^2 - a + 2*b)) + 2*(3*(a^5 - 3*a^4*b + 3*a
^3*b^2 - a^2*b^3)*tan(x)^5 + 3*(2*a^4*b - 9*a^3*b^2 + 11*a^2*b^3 - 4*a*b^4)*tan(x)^3 + (3*a^3*b^2 - 17*a^2*b^3
+ 22*a*b^4 - 8*b^5)*tan(x))*sqrt((a*tan(x)^2 + b)/tan(x)^2))/(a^6*b^2 - 3*a^5*b^3 + 3*a^4*b^4 - a^3*b^5 + (a^
8 - 3*a^7*b + 3*a^6*b^2 - a^5*b^3)*tan(x)^4 + 2*(a^7*b - 3*a^6*b^2 + 3*a^5*b^3 - a^4*b^4)*tan(x)^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{2}{\left (x \right )}}{\left (a + b \cot ^{2}{\left (x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)**2/(a+b*cot(x)**2)**(5/2),x)

[Out]

Integral(tan(x)**2/(a + b*cot(x)**2)**(5/2), x)

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Giac [B]  time = 1.40929, size = 351, normalized size = 2.49 \begin{align*} \frac{{\left (3 \, a^{3} \sqrt{b} \arctan \left (\frac{\sqrt{b}}{\sqrt{a - b}}\right ) - 3 \, \sqrt{a - b} a^{2} b + 14 \, \sqrt{a - b} a b^{2} - 8 \, \sqrt{a - b} b^{3}\right )} \mathrm{sgn}\left (\tan \left (x\right )\right )}{3 \,{\left (\sqrt{a - b} a^{5} \sqrt{b} - 2 \, \sqrt{a - b} a^{4} b^{\frac{3}{2}} + \sqrt{a - b} a^{3} b^{\frac{5}{2}}\right )}} - \frac{\arctan \left (\frac{\sqrt{a \tan \left (x\right )^{2} + b}}{\sqrt{a - b}}\right )}{{\left (a^{2} \mathrm{sgn}\left (\tan \left (x\right )\right ) - 2 \, a b \mathrm{sgn}\left (\tan \left (x\right )\right ) + b^{2} \mathrm{sgn}\left (\tan \left (x\right )\right )\right )} \sqrt{a - b}} - \frac{9 \,{\left (a \tan \left (x\right )^{2} + b\right )} a b^{2} - 6 \,{\left (a \tan \left (x\right )^{2} + b\right )} b^{3} - a b^{3} + b^{4}}{3 \,{\left (a^{5} \mathrm{sgn}\left (\tan \left (x\right )\right ) - 2 \, a^{4} b \mathrm{sgn}\left (\tan \left (x\right )\right ) + a^{3} b^{2} \mathrm{sgn}\left (\tan \left (x\right )\right )\right )}{\left (a \tan \left (x\right )^{2} + b\right )}^{\frac{3}{2}}} + \frac{\sqrt{a \tan \left (x\right )^{2} + b}}{a^{3} \mathrm{sgn}\left (\tan \left (x\right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^2/(a+b*cot(x)^2)^(5/2),x, algorithm="giac")

[Out]

1/3*(3*a^3*sqrt(b)*arctan(sqrt(b)/sqrt(a - b)) - 3*sqrt(a - b)*a^2*b + 14*sqrt(a - b)*a*b^2 - 8*sqrt(a - b)*b^
3)*sgn(tan(x))/(sqrt(a - b)*a^5*sqrt(b) - 2*sqrt(a - b)*a^4*b^(3/2) + sqrt(a - b)*a^3*b^(5/2)) - arctan(sqrt(a
*tan(x)^2 + b)/sqrt(a - b))/((a^2*sgn(tan(x)) - 2*a*b*sgn(tan(x)) + b^2*sgn(tan(x)))*sqrt(a - b)) - 1/3*(9*(a*
tan(x)^2 + b)*a*b^2 - 6*(a*tan(x)^2 + b)*b^3 - a*b^3 + b^4)/((a^5*sgn(tan(x)) - 2*a^4*b*sgn(tan(x)) + a^3*b^2*
sgn(tan(x)))*(a*tan(x)^2 + b)^(3/2)) + sqrt(a*tan(x)^2 + b)/(a^3*sgn(tan(x)))